Problem: $\dfrac{d}{dx}\left[\cos(x)x^2\right]=$
Explanation: $\cos(x)x^2$ is the product of two, more basic, expressions: $\cos(x)$ and $x^2$. Therefore, the derivative of the expression can be found using the product rule : $\begin{aligned} \dfrac{d}{dx}[u(x)v(x)]&=\dfrac{d}{dx}[u(x)]v(x)+u(x)\dfrac{d}{dx}[v(x)] \\\\ &=u'(x)v(x)+u(x)v'(x) \end{aligned}$ $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left(\cos(x)x^2\right) \\\\ &=\dfrac{d}{dx}\left[\cos(x)\right]x^2+\cos(x)\dfrac{d}{dx}(x^2)&&\gray{\text{The product rule}} \\\\ &=-\sin(x)x^2+\cos(x)2x&&\gray{\text{Differentiate }\cos(x)\text{ and }x^2} \\\\ &=2x\cos(x)-x^2\sin(x)&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $\dfrac{d}{dx}\left[\cos(x)x^2\right]=2x\cos(x)-x^2\sin(x)$ or any other equivalent form.